# When does a homotopy exist?

Let $$X\coprod X$$ be the disjoint union of $$X$$ with itself and let there be a commutative square

$$\require{AMScd} \begin{CD} X \coprod X @>{f,g}>> A \\ @V{(i_0, i_1)}VV @VV{p}V \\ X \times I @>>{k}> B \end{CD}$$

where $$p$$ is a Serre fibration and weak homotopy equivalence. Is it true that this diagram admits a lift $$h: X \rightarrow A$$ ? (In other words, is $$X \times I$$ a cylinder object for any $$X$$?). I know this is true for CW-complexes, but I think it holds for any topological space.

• I believe the only fact about CW complexes you would use in proving this is that the inclusion into the cylinder is a cofibration, but this is true for all spaces since it is clear that $X \times \{0\} \cup X \times \{1\}$ is a NDR of $X \times I$. – Connor Malin Sep 16 at 18:41
• I wrote something wrong, a need that both the top triangle and bottom triagle commute, in what a read about what you said implies the existence of lift making to top commute, but can we take such homotopy to make the bottom traingle commute? – HelloDarkness Sep 16 at 19:04
• Let $A$ be any space and take $p$ to be the unique map to a point. Then $f,g$ can be arbitrary maps and there is no reason why they should be homotopic. Clearly in general there is no reason why a diagonal should exist. – Tyrone Sep 16 at 19:06
• In this case $p$ wouldn't be serre fibration. – HelloDarkness Sep 16 at 19:07
• Yes, it would be. It would even be a Hurewicz fibration. – Tyrone Sep 16 at 19:08